Explanation:
The box is accelerating along the y-axis at a rate of [tex]+2.5\:\text{m/s}^2[/tex] as well as along the x-axis at a rate of [tex]+5.1\:\text{m/s}^2.[/tex] So the magnitude of the box's total acceleration is given by
[tex]a_T = \sqrt{a_x^2 + a_y^2}[/tex]
[tex]\:\:\:\:= \sqrt{(5.1\:\text{m/s}^2)^2 + (2.5\:\text{m/s}^2)^2}[/tex]
[tex]\:\:\:\:=5.7\:\text{m/s}^2[/tex]
The direction of the acceleration [tex]\theta[/tex] with respect to the horizontal direction is given by
[tex]\theta = \tan^{-1}\!\left(\dfrac{a_y}{a_x}\right) = \tan^{-1}\!\left(\dfrac{2.5\:\text{m/s}^2}{5.1\:\text{m/s}^2}\right)[/tex]
[tex]\:\:\:\:= 26.1°[/tex]
